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a: \(\dfrac{x}{x+3}=\dfrac{2}{5}\)
=>5x=2x+6
=>3x=6
hay x=2
b: \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{9}\)
=>x+1/2=2/3 hoặc x+1/2=-2/3
=>x=1/6 hoặc x=-7/6
c: \(\left(x-\dfrac{1}{2}\right)^3+\dfrac{1}{4}=\dfrac{3}{8}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{3}{8}-\dfrac{2}{8}=\dfrac{1}{8}\)
=>x-1/2=1/2
hay x=1
d: \(2^{x+3}=32\)
=>x+3=5
hay x=2
e: \(\Leftrightarrow3^x\left(3^2+1\right)=90\)
\(\Leftrightarrow3^x=9\)
hay x=2
b) \(3.2^{x+1}=12\)
\(2^{x+1}=12:3\)
\(2^{x+1}=4\)
\(2^{x+1}=2^2\)
\(x+1=2\)
\(x=2-1\)
\(x=1\)
Vậy \(x=1\)
c) \(2^{x-1}=2^3+2^4-2^3\)
\(2^{x-1}=8+16-8\)
\(2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(x-1=4\)
\(x=5\)
Vậy \(x=5\)
d) \(x^{50}=x\)
\(x^{50}-x=0\)
\(\Rightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
\(b.3.2^{x+1}=12\\ \Rightarrow2^{x+1}=4\\ \Rightarrow2^{x+1}=2^2\\ \Rightarrow x=1\\ \)
c) \(2^{x-1}=2^3-2^3+2^4\\ \Rightarrow2^{x-1}=0+16\\ \Rightarrow2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
d) \(x^{50}=x\\ \Rightarrow x=0;1\)
e) \(2\left(2x-1\right)^4=32\\ \Rightarrow\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\frac{3}{2}\)
g) Bí
a ) \(x\left(5-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right..\)
Vậy .....
b ) \(\left(3-x\right)\left(x^2+1\right)=0\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2=-1\left(loại\right)\end{matrix}\right.\)
Vậy ........
c ) giống câu b
d ) \(\left(2x-1\right)^3=8\Leftrightarrow2x-1=2\Leftrightarrow\Leftrightarrow x=\dfrac{3}{2}\)
e ) \(\left(x+3\right)^2=16\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)
Vậy.
b, \(\left(x+\frac{1}{2}\right)^2-\frac{1}{3}=\frac{1}{9}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\frac{4}{9}\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{3}\right)^3\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\Rightarrow x=\frac{1}{6}\)
c, \(\left(x-\frac{1}{2}\right)^3+\frac{1}{4}=\frac{3}{8}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{8}\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{2}\\x-\frac{1}{2}=-\frac{1}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)
d, \(2^{x+3}=32\)
\(\Leftrightarrow2^{x+3}=2^5\Leftrightarrow x+3=5\Rightarrow x=2\)
e, \(3^{x+2}+3^x=90\)
\(\Rightarrow\)\(3^x.\left(3^2+1\right)=90\)
\(\Rightarrow\)\(3^x.10=90\Rightarrow3^x=9\Rightarrow3^x=3^2\Rightarrow x=2\)